A243589 Numbers returned when each binary digit of n is replaced by the sum modulo 2 of the digits to its (wrapped) left and (wrapped) right.

0, 0, 0, 3, 5, 6, 0, 5, 15, 0, 10, 15, 5, 10, 0, 9, 27, 12, 30, 3, 17, 6, 20, 29, 15, 24, 10, 23, 5, 18, 0, 17, 51, 20, 54, 27, 57, 30, 60, 5, 39, 0, 34, 15, 45, 10, 40, 57, 27, 60, 30, 51, 17, 54, 20, 45, 15, 40, 10, 39, 5, 34, 0, 33, 99, 36, 102, 43, 105, 46

Offset: 1

Author

Anthony Sand, Jun 07 2014

Keywords

• nonn
• easy
• base
• look

Comments

a(n) = ror(n) XOR rol(n), where ror(x)=A038572(x) is x rotated one binary place to the right, rol(x)=A006257(x) is x rotated one binary place to the left, and XOR is the binary exclusive-or operator. - Alex Ratushnyak, May 24 2016

Numbers returned by the following function: take the t binary digits of n, d(1)..d(t), and replace each with the sum d(i) = (d(i-1) + d(i+1)) mod 2, where (i-1 = 0) maps to t and (i+1 > t) maps to 1.

Examples

			For 1, the function returns d(1) = (d(1) + d(1)) mod 2 = (1 + 1) mod 2 = 0.
For 5, the initial digits are (1,0,1).
d(1) = (d(3) + d(2)) mod 2 = (1 + 0) mod 2 = 1; d(2) = (d(1) + d(3)) mod 2 = (1 + 1) mod 2 = 0; d(3) = (d(2) + d(1)) mod 2 = (0 + 1) mod 2 = 1.
The function returns (1,0,1) = 101 = 5 in base 10.
		

Links

• Anthony Sand, Table of n, a(n) for n = 1..1000

Crossrefs

Cf. A006257, A035327, A038572, A055120.

Programs

Python

for n in range(1, 100):
    BL = len(bin(n))-2
    x = (n>>1) + ((n&1) << (BL-1))   # A038572(n)
    x^= (n*2) - (1<A006257(n)  for n>0
    print(x, end=', ')

Formula

for digits d(1)..d(t), d(i) = (d(i-1) + d(i+1)) mod 2, where (i-1 = 0) -> t, (i+1 > t) -> 1.